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Ivenika [448]
3 years ago
8

What is the answer to Sec (3.14/4) = ?

Mathematics
1 answer:
victus00 [196]3 years ago
7 0
0.785 would be your answer to 3.14/4

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The radius of a circle is 9.7 m. Find the circumference to the nearest tenth?(help its urgent)
yulyashka [42]

Answer:

Step-by-step explanation:

Circumference C=2πr  r=radius \pi =3.14

C=2(3.14)9.7

C=60.9

7 0
3 years ago
How do I find the inverse of this problem?
raketka [301]

\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Inverse of real function.

Let's consider here, g(n) = y ,

so we get as,

y =  \sqrt[3]{ \frac{n - 1}{2} }

no, cubing the power both side we get as,

=>

{y}^{3}  =  \frac{n - 1}{2}

now,

2 {y}^{3}  = n - 1

so finally, we get as,

=>

n = 2 {y}^{3}  + 1

hence,here, n = inverse of the g(n) function.

so,

g^-1 (n) = 2y^3+1.

7 0
3 years ago
5n + 3 > 4 - (6 – 5n)
jek_recluse [69]
3>-2 i believe
djdjjdksksk sorry i needed more characters
4 0
3 years ago
HURRY UP I NEED THE ANSWER
vladimir1956 [14]

Answer:

g(2) = 2

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given that g(x) = \frac{-1}{2} x + 3

Put x =2

               g(2) = \frac{-1}{2} (2) + 3

                       = -1+3

                       = 2

               g(2) = 2

5 0
3 years ago
What is the exact value of cos (67.5°)?
babymother [125]

first off, make sure you have a Unit Circle, if you don't do get one, you'll need it, you can find many online.

let's double up 67.5°, that way we can use the half-angle identity for the cosine of it, so hmmm twice 67.5 is simply 135°, keeping in mind that 135° is really 90° + 45°, and that whilst 135° is on the 2nd Quadrant and its cosine is negative 67.5° is on the 1st Quadrant where cosine is positive, so

cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\\\ cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}} \\\\[-0.35em] ~\dotfill\\\\ cos(135^o)\implies cos(90^o+45^o)\implies cos(90^o)cos(45^o)~~ - ~~sin(90^o)sin(45^o) \\\\\\ \left( 0 \right)\left( \cfrac{\sqrt{2}}{2} \right)~~ - ~~\left( 1\right)\left( \cfrac{\sqrt{2}}{2} \right)\implies -\cfrac{\sqrt{2}}{2} \\\\[-0.35em] ~\dotfill

cos(67.5^o)\implies cos\left( \frac{135^o}{2} \right)\implies \pm \sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}\implies \stackrel{I~Quadrant}{+\sqrt{\cfrac{ ~~ 1-\frac{\sqrt{2} ~~ }{2}}{2}}} \\\\\\ \sqrt{\cfrac{ ~~ \frac{2-\sqrt{2}}{2} ~~ }{2}}\implies \sqrt{\cfrac{2-\sqrt{2}}{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}\implies \cfrac{\sqrt{2-\sqrt{2}}}{2}

8 0
2 years ago
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